If you are not understand problem: when print undefined and when error, try to brute force it:)

p
n
p=0 and n=0    undefined
p=0 and n!=0   error
p!=0 and n=0   error
v (p=0 error, n=0 undefined, vg<=0 error)
t (n=0 error, p=0 undefined, tg<=0 error)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <stdio.h>
#include <cmath>

using namespace std;

int p=0,n=0,V=0,T=0;
char c;
int a=0,b=0,c1=0,d=0;
double R=8.314;

int main()
{
    for(int i = 0; i < 3; i++)
    {
        cin >> c;
        if(c=='p')
        {
            cin >> c;
            cin >> p;
            a++;
        }
        if(c=='n')
        {
            cin >> c;
            cin >> n;
            b++;
        }
        if(c=='V')
        {
            cin >> c;
            cin >> V;
            c1++;
        }
        if(c=='T')
        {
            cin >> c;
            cin >> T;
            d++;
        }

    }
    if(a==0) cout << "p = " << R*n*T/V;
    if(b==0) cout << "n = " << p*V/(R*T);
    if(c1==0)
    {
        if(p==0 && n==0)
        {
            cout << "undefined";
            return 0;
        }
        if(p==0 || n==0)
        {
            cout << "error";
            return 0;
        }
        cout << "V = " << R*n*T/p;
    }
    if(d==0)
    {
        if(p==0 && n==0)
        {
            cout << "undefined";
            return 0;
        }
        if(p==0 || n==0)
        {
            cout << "error";
            return 0;
        }
        cout << "T = " << p*V/(R*n);
    }
    return 0;
}

Так же было и на самом контесте WA 11, сейчас тоже WA 11, хотя на разборе говорили все тоже самое

Edited by author 20.03.2011 13:07

Edited by author 20.03.2011 13:07

in 12 test answer is p = 0.
I write in my code "0" instead of "p = 0" and have WA12.
Good Luck!

double res = (bpi*bvi)/(bti*R);
cout << "n = ";
if(res < 1)
    printf("%.0f", res);
else
    printf("%.6f", res);

When res (n) < 1 u should cout 0 else u should cout number with an error.
Good Luck!

Can anyone help me with test 4?

#include <iostream>
#include <cstdio>

using namespace std;

int main() {
    char c[3];
    float v[3];
    float R = 8.314, p, V, T, n;
    scanf("%c = %f\n", &c[0], &v[0]);
    scanf("%c = %f\n", &c[1], &v[1]);
    scanf("%c = %f", &c[2], &v[2]);
    for(int i = 0; i < 3; i++) {
        if(c[i] == 'p') p = v[i];
        else if(c[i] == 'V') V = v[i];
        else if(c[i] == 'T') T = v[i];
        else if(c[i] == 'n') n = v[i];
    }
    if(c[0] != 'n' && c[1] != 'n' && c[2] != 'n') {
        n = p*V/(R*T);
        if(T == 0 && (p == 0 || V == 0)) cout << "undefined\n";
        else if(T == 0 && (p != 0 || V != 0)) cout << "error\n";
        else if(p < 0 || V < 0 || T < 0 || n < 0) cout << "error\n";
        else printf("n = %f\n", n);
    }
    else if(c[0] != 'T' && c[1] != 'T' && c[2] != 'T') {
        T = p*V/(R*n);
        if(n == 0 && (p == 0 || V == 0)) cout << "undefined\n";
        else if(n == 0 && (p != 0 || V != 0)) cout << "error\n";
        else if(p < 0 || V < 0 || n < 0 || T < 0) cout << "error\n";
        else printf("T = %f\n", T);
    }
    else if(c[0] != 'V' && c[1] != 'V' && c[2] != 'V') {
        V = n*T*R/p;
        if(p == 0 && (T == 0 || n == 0)) cout << "undefined\n";
        else if(p == 0 && (T != 0 || n != 0)) cout << "error\n";
        else if(p < 0 || T < 0 || n < 0 || V < 0) cout << "error\n";
        else printf("V = %f\n", V);
    }
    else if(c[0] != 'p' && c[1] != 'p' && c[2] != 'p') {
        p = n*T*R/V;
        if(T == 0 && (n == 0 || V == 0)) cout << "undefined\n";
        else if(T == 0 && (n != 0 || V != 0)) cout << "error\n";
        else if(V < 0 || n < 0 || T < 0) cout << "error\n";
        else printf("p = %f\n", p);
    }
    return 0;
}

What is test 6? I am stuck at it for a very long time!

if ((p = 0) and (n = 0)) and ((Pc = 1) and (Nc = 1)) then
  writeln('undefined');

if (((p = 0) and (n <> 0)) or ((p <> 0) and (n = 0)))and ((Pc = 1) and (Nc = 1)) then
  writeln('error');

this pair if we have on input n and p (Nc = 1 and Pc = 1)
but still i have wa on 6 test!
are there any conditions for "error" and "undefined"??? please help. i've already sent my solution for 13 times! and still it's wa...

Hi.
I find out that test #3 looks like this :
p = 3
V = 4
n = 0

When we need to find T, answer will be "undefined" if and only if n = 0 (because, in this way, we get division by zero).
In second and in third tests n=0, and we need to find T. I wrote code like this :

if( we need find T ) {
    if( n==0 )
         printf("undefined\n");
}


But, in this way, I get WA 3.
What is wrong? Please, check test, or, if I made mistake, help me figure it out.
Thanks.

Edited by author 16.07.2012 01:48

When n = 0,
if we have to calculate T, is the answer 'undefined' ?
if we have to calculate V, is the answer 'error' or 'undefined'?

if n or p or T or V not integer then error or some value?

Edited by author 15.01.2012 05:32

error  if :
        ( IsUsed[1] && V == 0 ) ||
        ( IsUsed[3] && T == 0 ) ||
        ( IsUsed[0] && IsUsed[2] && p*n==0 && ( p!=0 || n!=0 ) )

undefined if:
        ( IsUsed[0] && IsUsed[2] && p==0 && n==0 )

It's so hard! =)

Print "p = ", but not "P = "
And "n = ", but not "N = "

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;


public class T_1823 {

    public static final double R = 8.314;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        Map<Character, Double> mp = new HashMap<Character, Double>();
        for (int i = 1; i <=3; i++) {
            String s = in.nextLine().trim();
            mp.put(s.charAt(0), (double)Integer.parseInt(s.charAt(4)+""));
        }
        if(mp.get('T')!=null) {
            if(mp.get('T')==0) {
                System.out.println("error");
                return;
            }
        }
        if(mp.get('V')!=null) {
            if(mp.get('V')==0) {
                System.out.println("error");
                return;
            }
        }
        //System.out.println(mp);
        if(!mp.containsKey('p')) {
            double p = mp.get('n')*mp.get('T')*R/mp.get('V');
            System.out.print("p = ");
            if(p==0) System.out.println(0);
            else System.out.printf("%.6f",p);
        }
        if(!mp.containsKey('n')) {
            double n = mp.get('p')*mp.get('V')/R*mp.get('T');
            System.out.print("n = ");
            if(n==0) System.out.print(0);
            else System.out.printf("%.6f",n);
        }
        if(!mp.containsKey('T')) {
            if(mp.get('n')==0 && mp.get('p')!=0) {
                System.out.println("error");
            } else {
                if(mp.get('n')==0 && mp.get('p')==0) {
                    System.out.println("undefined");
                } else {
                    double T = mp.get('p')*mp.get('V')/R*mp.get('n');
                    if(T==0) {
                        System.out.println("error");
                    } else {
                        System.out.print("T = ");
                        System.out.printf("%.6f",T);
                    }
                }
            }
        }
        if(!mp.containsKey('V')) {
            if(mp.get('p')==0 && mp.get('n')!=0) {
                System.out.println("error");
            } else {
                if(mp.get('n')==0 && mp.get('p')==0) {
                    System.out.println("undefined");
                } else {
                    double V = mp.get('n')*R*mp.get('T')/mp.get('p');
                    if(V==0) {
                        System.out.println("error");
                    } else {
                        System.out.print("V = ");
                        System.out.printf("%.6f",V);
                    }
                }
            }
        }
    }
}

Help ! What is test case #2 ?

V or T is calculated to be 0 => output error

Please give test.
(I already use printf instead of cout <<)

Edited by author 29.06.2011 21:58

Edited by author 29.06.2011 21:58

p = 1
t = 0
v = 1


i have AC but WA on this test

What is test 3?
Can't find any mistakes, but WA#3.